| Exam Board | OCR |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2011 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Straight Lines & Coordinate Geometry |
| Type | Gradient from equation or points |
| Difficulty | Moderate -0.8 This is a straightforward multi-part question testing basic coordinate geometry: distance formula, gradient calculation, and perpendicular line condition. All parts are routine applications of standard formulas with no problem-solving insight required, making it easier than average but not trivial since it requires multiple techniques. |
| Spec | 1.03b Straight lines: parallel and perpendicular relationships1.10f Distance between points: using position vectors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\sqrt{(-2-6)^2 + (7-1)^2}\) | M1 | Use of \(\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\); 3 out of 4 substitutions correct; Look out for no square root, \((x_2+x_1)^2\) etc. M0 |
| \(= 10\) | A1 | 2 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{7-1}{-2-6}\) | M1 | Uses \(\frac{y_2-y_1}{x_2-x_1}\); 3 out of 4 substitutions correct |
| \(= -\frac{3}{4}\) | A1 | 2 marks; o.e. ISW; Allow \(-0.75\), \(\frac{3}{-4}\) etc. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Gradient of given line \(= \frac{4}{3}\) | M1 | Attempt to rearrange equation to make \(y\) the subject OR attempt to find gradient using points on line; Must at least isolate \(y\) |
| \(-\frac{3}{4} \times \frac{4}{3} = -1\) | B1ft | Correct conclusion for their gradients |
| So lines are perpendicular | B1 | 3 marks (total 7); States \(-\frac{3}{4} \times \frac{4}{3} = -1\) or "negative reciprocal" relating to correct values www |
# Question 1:
## Part (i)
| Answer | Mark | Guidance |
|--------|------|----------|
| $\sqrt{(-2-6)^2 + (7-1)^2}$ | M1 | Use of $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$; 3 out of 4 substitutions correct; Look out for no square root, $(x_2+x_1)^2$ etc. **M0** |
| $= 10$ | A1 | **2 marks** |
## Part (ii)
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{7-1}{-2-6}$ | M1 | Uses $\frac{y_2-y_1}{x_2-x_1}$; 3 out of 4 substitutions correct |
| $= -\frac{3}{4}$ | A1 | **2 marks**; o.e. **ISW**; Allow $-0.75$, $\frac{3}{-4}$ etc. |
## Part (iii)
| Answer | Mark | Guidance |
|--------|------|----------|
| Gradient of given line $= \frac{4}{3}$ | M1 | Attempt to rearrange equation to make $y$ the subject **OR** attempt to find gradient using points on line; Must at least isolate $y$ |
| $-\frac{3}{4} \times \frac{4}{3} = -1$ | B1ft | Correct conclusion for their gradients |
| So lines are perpendicular | B1 | **3 marks** (total **7**); States $-\frac{3}{4} \times \frac{4}{3} = -1$ or "negative reciprocal" relating to correct values **www** |
---1 The points $A$ and $B$ have coordinates $( 6,1 )$ and $( - 2,7 )$ respectively.\\
(i) Find the length of $A B$.\\
(ii) Find the gradient of the line $A B$.\\
(iii) Determine whether the line $4 x - 3 y - 10 = 0$ is perpendicular to $A B$.
\hfill \mbox{\textit{OCR C1 2011 Q1 [7]}}